Question

Find the solution to the given systems

(a) X'=$\begin{bmatrix} 1/2 &0 \\ 1& -1/2 \end{bmatrix}$X X(0)=$\bigl(\begin{smallmatrix} 3\\ 5 \end{smallmatrix}\bigr)$

(b) dx/dt=5x+y dy/dt=-2x+3y

Answer 1

given X' Xlo) +112 Here A = 12 12 Lel Abe an eigenvalue of A then det (A-AI)=0 72-7 = (3-1)(31-4) = 0 -1 2-2 7 = ² 1 - 1 / 2 ^, = +22= 2 are eigenvalue of A. let be eigenvector of A corresponding to A, ~ 122 AX, = 2 X (1–3 1 )X, = 0 1 2 2 - 2 2 2 Hence X then O = 0 ) - -1-1 2 2 1 N2 N2 02 22 let hy - 12 20 n = n2 12 [ X = 0 is an eigenvector of corresponding to ^, = 1 / 2 an eigenvector of A corresponding to 82-1 then А Let 12 be

12 = -1 then (A + 2 IX [***]]-1 no Mence X2 = 312 82 72 let na = 1 is an eigenvector of A is then X2 = (0) A corresponding to I2 = -² Therefore, solution of equation X (t) C, Ex + C e where at C2 tra 3 X (2) = C, 8th (1) + (2 872 (8) and X2 are constant *12 X (t) = Cle 412 412 ce + Cae we get using X (0) = ( 3 ) X60) = 0 3 (3) lente) & C+C2=5 - C2 = 5-3=2 412 3 e IX (1) ho $2+2e is required solution 3e

- Given dx Sxty f dy at = -2X+34 - dt Differentiating o o.sito 'l get x = sx' ty > X = Sx' + (-2X+34) using y'= -2X+34 from ② ) = xl - SX'-28 +34 (from 0 y = xl-5x) SX' - 2x + 3 + 3(xl-5x) 8x' - 178 X - 8x'+17% =0 X!! Xll The characteristic equation of is z 288 +17 20 8 8 164-68 2 8 121 44 2 This is of the form 8= VIPi where = 4 + Bry Therefore solution of ② is - X(t) = e (c, cosplt cz 44 singt) X (2) (c, cost & Ca sint) = x '(t) = 4 et (co cast + Ca sint ) + tlf-Gsint & Ca cost) e we using & f in 0 get

44 tett 4 et Cc, cost & Casint) (-c, sind + C₂ cost) = sett (a cosa tez sint ) + x Y = -cte (ci cose + C2 sind) + All-C, sint + C2 cosf) at Ja y (2) = -c, ett (costa sint) tet & C₂ ( cost-sint) Hence, equation of are the required solution At